Question: Let $g(x)=\dfrac{1}{4}x^4-4x^3+24x^2$. For what values of $x$ does the graph of $g$ have a point of inflection? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=-3$ (Choice B) B $x=0$ (Choice C) C $x=4$ (Choice D) D $g$ has no points of inflection.
Explanation: We can find the inflection points of the graph of $g$ by looking for the intervals where its second derivative $g''$ is positive/negative. This analysis is very similar to finding minimum/maximum points, only instead of analyzing $g'$, we are analyzing $g''$. The second derivative of $g$ is $g''(x)=3(x-4)^2$. $g''(x)=0$ for $x=4$. Since $g''$ is a polynomial, it's defined for all real numbers. Therefore, our possible inflection point is $x=4$. Our possible inflection points divide the number line into two intervals: $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$ $8$ $(-\infty,4)$ $(4,\infty)$ Let's evaluate $g''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $g''(x)$ Verdict $(-\infty,4)$ $x=3$ $g''(3)=3>0$ $g$ is concave up $\cup$ $(4,\infty)$ $x=5$ $g''(5)=3>0$ $g$ is concave up $\cup$ We can see that the graph of $g$ does not change concavity at $x=4$. Therefore, $x=4$ is not a point of inflection. In conclusion, $g$ has no points of inflection.